A050482 Sum of remainders when n-th prime is divided by all preceding integers.

0, 1, 4, 8, 22, 28, 51, 64, 98, 151, 167, 233, 297, 325, 403, 505, 635, 645, 790, 904, 923, 1113, 1244, 1422, 1654, 1800, 1888, 2056, 2098, 2256, 2849, 3066, 3326, 3450, 3969, 4045, 4329, 4696, 5014, 5325, 5767, 5759, 6499, 6565, 6898

Offset

1,3

Comments

a(n)/(n*log(n))^2 appears to approach a constant ~0.22... for large n. - Benedict W. J. Irwin, Dec 07 2016

Irwin's comment is incorrect. - Bill McEachen, Feb 04 2024. [Indeed, according to the first formula in A004125, a(n)/(n*log(n))^2 approaches a constant, which is not 0.22 but 1-Pi^2/12 = 0.1775... - Amiram Eldar, Feb 04 2024]

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Formula

a(n) = A004125(A000040(n)). - R. J. Mathar, Jun 12 2009

Example

a(4) = 8 because remainders when 7 is divided by 1..6 are 0,1,1,3,2,1, which add to 8.
a(2) = 3 mod (3-1) = 1.
a(3) = (5 mod (5-1)) + (5 mod (5-2)) + (5 mod (5-3)) = 2 + 1 + 1 = 4.

Maple

A050482 := proc(n) local a, i; a := 0; for i from 1 to ithprime(n)-1 do a := a+(ithprime(n) mod i); od: end;

Mathematica

Table[Sum[Mod[Prime[n], k], {k, Prime[n]-1}], {n, 45}] (* James C. McMahon, Feb 08 2024 *)

Prog

(PARI) a(n)=my(p=prime(n)); sum(k=2, p, p%k) \\ Charles R Greathouse IV, Jun 03 2013
(Python)
from math import isqrt
from sympy import prime
def A050482(n): return (p:=prime(n))**2+((s:=isqrt(p))**2*(s+1)-sum((q:=p//k)*((k<<1)+q+1) for k in range(1, s+1))>>1) # Chai Wah Wu, Nov 01 2023

Crossrefs

Cf. A004125, A034953.

Sequence in context: A129794 A370956 A064503 * A323584 A332199 A200149

Adjacent sequences: A050479 A050480 A050481 * A050483 A050484 A050485

Keyword

nonn

Author

Brian Wallace (wallacebrianedward(AT)yahoo.co.uk), Dec 26 1999

Status

approved